In triangle ABC, angle A is 60 degrees, AC / AB = (√3 + 1) / 2. Find the angle B in degrees.

Using the cosine theorem, we get:

BC ^ 2 = AB ^ 2 + AC ^ 2 – 2 * AB * AC * cos (A).

Since AC = (√3 + 1) / 2 * AB by the condition:

BC ^ 2 = AB ^ 2 + (√3 + 1) / 2) ^ 2 * AB ^ 2 – AB ^ 2 * (√3 + 1);

BC = AB * √ (4 + (√3 + 1) ^ 2) / 4 – (√3 + 1)).

Using the theorem of sines, we get the expression:

AC / sin (B) = BC / sin (A);

sin (B) = AC * sin (A) / BC;

sin (B) = (√3 + 1) * √3 / 2 * √ (4 + (√3 + 1) ^ 2) / 4 – (√3 + 1)).

B = arcsin ((√3 + 1) * √3 / 2 * √ (4 + (√3 + 1) ^ 2) / 4 – (√3 + 1))).